Circle A has a radius of 2  and a center at (3 ,6 ). Circle B has a radius of 5  and a center at (2 ,3 ). If circle B is translated by <-2 ,1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

May 16, 2016

circles overlap.

Explanation:

What we have to do here is compare the distance (d) between the centres of the circles to the sum of the radii.

• If sum of radii > d , then circles overlap

• If sum of radii < d , then no overlap

The first step is to find the new centre of B under the translation. A translation does not change the shape of a figure , only it's position.

Under a translation of $\left(\begin{matrix}- 2 \\ 1\end{matrix}\right)$

centre B(2 ,3) → (2-2 ,3+1) → B(0 ,4)-(new centre)

To calculate the distance (d) between the centres use the $\textcolor{b l u e}{\text{ distance formula}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

The 2 points here being the centres of A and B.

let $\left({x}_{1} , {y}_{1}\right) = \left(3 , 6\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(0 , 4\right)$

d=sqrt((0-3)^2+(4-6)^2)=sqrt(9+4)=sqrt13≈3.606

radius of A + radius of B = 2 + 5 = 7

Since sum of radii > d , then circles overlap.